Question

7. A 0.250-g sample of calcium metal is heated in a 38.500-g crucible to form calcium oxide. The resulting crucible and product weigh 38.850 g. Refer to Example Exercise 1 and determine the empirical formula of calcium oxide.

Answer 1

Ans 7 :

mass of calcium = 0.250 g

mol calcium = mass / molar mass

= 0.250 g / 40.078 g/mol = 0.00624 mol

mass of compound = mass of crucible and compound - mass of crucible

= 38.850 g - 38.500 g

= 0.350 g

mass of oxygen = mass of compound - mass calcium

= 0.350 g - 0.250 g

= 0.100 g

mol oxygen = 0.100 g / 15.999 g/mol = 0.00625 mol

dividing the number of mol by a common value we get the ratio of number of mol as 1 : 1

so the empirical formula of the compound will be : CaO